WAEC SSCE - General Mathematics - 2014

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The statement "F x varies inversely as y" can be mathematically written as x = k/y, where k is the constant of proportionality. Similarly, "y varies directly as Z" can be written as y = kz, where k is again the constant of proportionality. Substituting y in the first equation, we get x = k/(kz), which simplifies to x = 1/z. Therefore, we can conclude that x varies inversely with z, which is mathematically represented as x \(\alpha\) 1/z. So, the correct answer is: x \(\alpha \frac{1}{z}\).

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To solve this problem, we need to follow the given operations in order of precedence, i.e., first we have to find the complement of y, then union it with z and finally find the intersection of x with the resulting set. The complement of y (denoted by Y') is the set of all elements in u that are not in y. So, Y' = {0, 5, 6}. The union of Y' and Z (denoted by Y' \(\cup\) Z) is the set of all elements that are in Y' or Z or both. So, Y' \(\cup\) Z = {0, 1, 3, 5, 6}. Finally, the intersection of x with Y' \(\cup\) Z (denoted by x \(\cap\) (Y' \(\cup\) Z)) is the set of all elements that are in both x and Y' \(\cup\) Z. So, x \(\cap\) (Y' \(\cup\) Z) = {0, 6}. Therefore, the answer is {0, 6}.

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The sum of the exterior angles of a polygon is always 360 degrees. Therefore, if each exterior angle is 30 degrees, the polygon must have 360/30 = 12 sides. The formula to find the sum of interior angles of a polygon is (n-2) x 180 degrees, where n is the number of sides. Substituting n = 12 into the formula, we get: (12-2) x 180 = 10 x 180 = 1800 degrees Therefore, the sum of the interior angles of this polygon is 1800 degrees. So the correct option is 1800^o.

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To simplify the expression 10\(\frac{2}{5} - 6 \frac{2}{3} + 3\), we first need to convert the mixed numbers to improper fractions. 10\(\frac{2}{5}\) can be written as \(\frac{52}{5}\) and 6\(\frac{2}{3}\) can be written as \(\frac{20}{3}\). So, the expression becomes: \begin{align*} 10\frac{2}{5} - 6 \frac{2}{3} + 3 &= \frac{52}{5} - \frac{20}{3} + 3\\ &= \frac{156}{15} - \frac{100}{15} + \frac{45}{15}\\ &= \frac{101}{15} \end{align*} Therefore, the simplified expression is 6\(\frac{11}{15}\). Hence, the correct option is: \boxed{2. 6\frac{11}{15}}

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To approximate 0.0033780 to 3 significant figures, we need to count the first three non-zero digits, then round off the remaining digits. The first three non-zero digits are 3, 3, and 7, so we keep them. The next digit is 8, which is greater than or equal to 5, so we round up the last digit (0) to 1. Therefore, the answer is 0.00338 (option c).

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We have: \begin{align*} &\frac{1}{2}(a-b+c)+\frac{1}{2}(a+b-c)-\frac{1}{2}(a-b-c)\\ &=\frac{1}{2}(2a)\\ &=a \end{align*} Therefore, the expression simplifies to $a$. So, the answer is: $\frac{1}{2}(a+b+c)$.

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To find the gradient of the line joining two points, we use the formula: Gradient = (change in y) / (change in x) We can calculate the change in y by subtracting the y-coordinate of one point from the y-coordinate of the other point, and similarly, we can calculate the change in x by subtracting the x-coordinate of one point from the x-coordinate of the other point. In this case, the two points are (2, -3) and (2, 5). The change in y is: 5 - (-3) = 8 The change in x is: 2 - 2 = 0 Since the change in x is 0, we cannot divide by it, and the gradient is undefined. This means that the line joining these two points is vertical and has no slope. Therefore, the correct answer is: undefined.

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To find the coordinates of point S on the graph, we need to use the equation of the graph, which is given as y = 2x^2 + x - 1. From the graph, we can see that the point S has x-coordinate of 1. To find the y-coordinate, we can substitute x=1 into the equation and solve for y: y = 2(1)^2 + (1) - 1 y = 2 + 1 - 1 y = 2 Therefore, the coordinates of point S are (1, 2). So, the answer is (c) (1, 2.0).

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To find the probability that the sum of the numbers is prime, we need to first determine all the possible sums of numbers from sets P and Q. The elements in set P are {1, 2, 3}, and the elements in set Q are {2, 3, 5}. To find all the possible sums, we can create a table of sums by adding each element of set P to each element of set Q: | | 2 | 3 | 5 | |---|---|---|---| | 1 | 3 | 4 | 6 | | 2 | 4 | 5 | 7 | | 3 | 5 | 6 | 8 | From this table, we see that there are nine possible sums. To determine which of these sums are prime, we can simply check each one: - 3: Not prime - 4: Not prime - 5: Prime - 6: Not prime - 7: Prime - 8: Not prime - 9: Not prime - 10: Not prime - 11: Prime So, there are three possible sums that are prime: 5, 7, and 11. Therefore, the probability that the sum of the numbers is prime is: $$\frac{\text{number of favorable outcomes}}{\text{total number of possible outcomes}} = \frac{3}{9} = \frac{1}{3}$$ Therefore, the answer is option (C) \(\frac{1}{3}\).

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To calculate the mean height of the plants, we need to first find the sum of all the heights and divide it by the total number of plants. We can do this using the formula: mean = (sum of all heights) / (total number of plants) To find the sum of all the heights, we need to multiply each height by its corresponding frequency and add up the results. So, the sum of all heights = (2 x 2) + (3 x 4) + (4 x 5) + (5 x 3) + (6 x 1) = 4 + 12 + 20 + 15 + 6 = 57 The total number of plants is the sum of all the frequencies: 2 + 4 + 5 + 3 + 1 = 15. Therefore, the mean height of the plants = 57 / 15 = 3.8. Hence, the answer is 3.8.

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The formula we want to solve for u is E = (m/2g)(v^2 - u^2).

To make u the subject of the formula, we need to isolate u on one side of the equation by performing the necessary algebraic operations.

First, we'll simplify the right side of the equation:

E = (m/2g)(v^2 - u^2)
2gE/m = v^2 - u^2    // Multiply both sides by 2g/m

Next, we'll isolate u^2 by adding it to both sides of the equation:

2gE/m + u^2 = v^2

Finally, we'll solve for u by taking the square root of both sides of the equation:

u = sqrt(v^2 - 2gE/m)

Therefore, the value of u as the subject of the formula is:

u = sqrt(v^2 - 2gE/m)

Option A is the correct answer.

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Let's start by using the fact that PQR is a straight line, which means that m + n + r = 180°. We can then use the given information to form two equations and solve for m + r. From (m + n) = 120°, we can substitute n = 120° - m to get: (m + 120° - m + r) = 180° r + 120° = 180° r = 60° From (n + r) = 100°, we can substitute n = 120° - m and r = 60° to get: (120° - m + 60°) = 100° m = 80° Finally, we can find m + r by adding m and r: m + r = 80° + 60° = 140° Therefore, the answer is 140°.

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To find the value of x, we need to determine what base value the number 23_x represents. We are told that 23_x is equal to 32₅. In other words, the value represented by 23_x is equal to the value represented by 32₅. We can convert 32₅ to base 10 to get: 32₅ = (3 x 5¹) + (2 x 5⁰) = 15 + 2 = 17 So, we have: 23_x = 17 To find the value of x, we need to determine what base value raised to what power is equal to 17. We can do this by trying different values of x. If x is 2, then 23₂ = (2 x 2¹) + (3 x 2⁰) = 4 + 3 = 7 If x is 3, then 23₃ = (2 x 3¹) + (3 x 3⁰) = 6 + 3 = 9 If x is 4, then 23₄ = (2 x 4¹) + (3 x 4⁰) = 8 + 3 = 11 If x is 5, then 23₅ = (2 x 5¹) + (3 x 5⁰) = 10 + 3 = 13 If x is 6, then 23₆ = (2 x 6¹) + (3 x 6⁰) = 12 + 3 = 15 If x is 7, then 23₇ = (2 x 7¹) + (3 x 7⁰) = 14 + 3 = 17 So, we have found that if x is 7, then 23_x = 17, which matches the value of 32₅. Therefore, the value of x is 7. Answer: 7

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Let's draw a diagram to visualize the situation:

        N
        |
        F
        |
W-------+-------E
        |
        |
        S

The ship starts at the point marked "W", then sails east to reach the point marked "E". The distance between W and E is x km. Then the ship sails north from E to reach the point marked "F". The distance between E and F is also x km.

We want to find the bearing of F from W. This is the angle that the line segment WF makes with the north-south line, measured in a clockwise direction.

Let's call the point where the north-south line intersects the line WE as point G. Then we have a right triangle WGE, where WG is the distance travelled east by the ship, and GE is the distance travelled north. The angle WGE is the bearing we're looking for.

From the right triangle WGE, we can use trigonometry to find the angle WGE:

tan(WGE) = opposite / adjacent = GE / WG = x / x = 1

Taking the arctan of both sides, we get:

WGE = arctan(1) = 45 degrees

Therefore, the bearing of F from W is 045 degrees. Answer: (a) 045^o.

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We can use trigonometric identities to evaluate the expression. We know that: - $\tan x = \frac{\sin x}{\cos x}$. - $\sin^2x + \cos^2x = 1$. We can use the second identity to find $\sin x$: $$\begin{aligned} \sin^2x + \cos^2x &= 1 \\ \sin^2x &= 1 - \cos^2x \\ \sin x &= \sqrt{1 - \cos^2x} \\ &= \sqrt{1 - \left(\frac{12}{13}\right)^2} \\ &= \frac{5}{13} \end{aligned}$$ Now we can substitute $\sin x$ and $\cos x$ into the expression we need to evaluate: $$\begin{aligned} \frac{1 - \tan x}{\tan x} &= \frac{1 - \frac{\sin x}{\cos x}}{\frac{\sin x}{\cos x}} \\ &= \frac{\cos x - \sin x}{\sin x} \\ &= \frac{\frac{12}{13} - \frac{5}{13}}{\frac{5}{13}} \\ &= \frac{7}{5} \end{aligned}$$ Therefore, the answer is $\frac{7}{5}$.

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To find the length of a side of a cube, we need to take the cube root of its volume. Given that the volume of the cube is 512cm³, we have: side³ = volume side³ = 512cm³ Taking the cube root of both sides, we get: side = ∛(512cm³) side = 8cm Therefore, the length of the side of the cube is 8cm. Answer: Option (C) 8cm

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To find the minimum value of y for the given relation y = 2x^2 + x - 1, we need to locate the vertex of the parabolic graph. The vertex of a parabola with equation y = ax^2 + bx + c is given by the coordinates (-b/2a, c - b^2/4a). In this case, a = 2, b = 1, and c = -1. Substituting these values into the formula, we get: Vertex x-coordinate = -b/2a = -1/(2*2) = -1/4 Vertex y-coordinate = 2(-1/4)^2 + (1/4) - 1 = -1.25 Therefore, the minimum value of y is -1.25. So the correct option is (C) -1.25.

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To find the distance of the chord from the center of the circle, we can use the formula: distance from center = √(r^2 - (c/2)^2) where r is the radius of the circle and c is the length of the chord. In this case, r = 3.7 cm and c = 7 cm. Therefore: distance from center = √(3.7^2 - (7/2)^2) = √(13.69 - 12.25) = √1.44 = 1.2 cm Therefore, the distance of the chord from the center of the circle is 1.2 cm. The logic behind the formula is that the distance from the center of the circle to the chord is the perpendicular distance from the center to the line that contains the chord. If we draw radii from the center of the circle to the endpoints of the chord, we can form a right triangle with the chord as the hypotenuse. The distance from the center to the chord is the height of this triangle, which can be found using the Pythagorean theorem.

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A measure of dispersion describes how spread out or varied a set of data is. Among the options given, only the range is a measure of dispersion. Range is calculated by subtracting the minimum value from the maximum value in a dataset. It provides information about the spread of the data from the lowest to the highest value. Percentile, median, and quartile are measures of central tendency, which describe the typical or central value in a dataset.

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The volume of a cylinder is given by the formula V = πr²h, where r is the radius of the base of the cylinder and h is the height of the cylinder. The volume of the water in tin P is V_P = πr²h_P, where r is the radius of the base of tin P and h_P is the height of the water in tin P. The volume of the same quantity of water in tin Q is V_Q = π(2r)²h_Q = 4πr²h_Q, where 2r is the radius of the base of tin Q and h_Q is the height of the water in tin Q. Since the same quantity of water is in both tins, V_P = V_Q. Substituting the expression for V_P and V_Q into the equation V_P = V_Q gives πr²h_P = 4πr²h_Q, which simplifies to h_Q = h_P/4. Therefore, the height of the water in tin Q is one-fourth of the height of the water in tin P. Since the height of the water in tin P is 10cm, the height of the water in tin Q would be (10/4) cm = 2.5 cm. Therefore, the height of the same quantity of water in tin Q is 2.5 cm. Answer option (A) is the correct answer.

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The area of a sector of a circle is given by the formula: A = (θ/360)πr^2 where A is the area of the sector, θ is the angle of the sector in degrees, r is the radius of the circle. Let's first find the radius of the circle. The diameter of the circle is given as 12 cm. Therefore, the radius is half of the diameter, i.e., r = 12/2 = 6 cm. The area of the sector is given as 66 cm^2. Therefore, 66 = (θ/360)π(6^2) 66 = (θ/360)π(36) 66 = (θ/10)π θ = (660/π) × 10/360 θ = 18.86° (approx.) Now, we can use the radius of the circle and the angle of the sector to find the slant height (l) of the cone: l = √(r^2 + h^2), where h is the height of the cone. The lateral area of the cone (i.e., the curved surface area) is half the area of the sector: L = (1/2)A = (1/2) × 66 = 33 cm^2 The slant height of the cone is given by: l = L/r = 33/6 = 5.5 cm We can use the Pythagorean theorem to find the height of the cone: h = √(l^2 - r^2) = √(5.5^2 - 6^2) = √(30.25 - 36) = √5.75 Therefore, the height of the cone is h = 2.4 cm (approx.). Finally, we can use the formula for the volume of a cone to find the radius of the base: V = (1/3)πr^2h We know the volume of the cone is equal to the area of the sector: V = A = (θ/360)πr^2 = (18.86/360) × π × r^2 Therefore, (1/3)πr^2h = (18.86/360) × π × r^2 h = (18.86/360) × 3 r^2 = (3/π) × (360/18.86) × 66 r = √((3/π) × (360/18.86) × 66) = 3.5 cm (approx.) Hence, the radius of the base of the cone is 3.5 cm. Therefore, the correct option is (b) 3.5cm.

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Let's assume that there are x N50 notes and y N20 notes in the box. From the problem, we know that the total number of currency notes is 13, so we have: x + y = 13 ----(1) We also know that the total value of the currency notes is N530. If we express this in terms of the number of N50 and N20 notes, we get: 50x + 20y = 530 ----(2) We now have two equations (1) and (2) with two variables (x and y), which we can solve simultaneously. Multiplying equation (1) by 20, we get: 20x + 20y = 260 Subtracting this equation from equation (2), we get: 30x = 270 Dividing both sides by 30, we get: x = 9 Therefore, there are 9 N50 notes in the box. Answer: 9

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Let's assume the number we are looking for to be 'x'. According to the problem statement, three quarters of the number can be written as \(\frac{3}{4}x\) and two and a half of the number can be written as \(2.5x\). It is given that the sum of these two quantities gives 13. So we can write it as: \(\frac{3}{4}x + 2.5x = 13\) Now, we can simplify this equation by combining the two terms on the left-hand side: \(\frac{3}{4}x + 2.5x = 13\) \(\Rightarrow \frac{3}{4}x + \frac{10}{4}x = 13\) \(\Rightarrow \frac{13}{4}x = 13\) \(\Rightarrow x = \frac{13 \times 4}{13}\) \(\Rightarrow x = 4\) Therefore, the number we are looking for is 4. Hence, the answer is 4.

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(a)(i) \(f(x) = 2x^2 - 1\), so \(f(x-1) = 2(x-1)^2 - 1\).

\[2(x-1)^2 - 1 - 7 = 0 \;\Rightarrow\; 2(x-1)^2 = 8 \;\Rightarrow\; (x-1)^2 = 4\]

\[x - 1 = \pm 2 \;\Rightarrow\; x = 3 \text{ or } x = -1\]

(ii) Evaluate each term:

• \(f\!\left(-\tfrac12\right) = 2\left(\tfrac14\right) - 1 = -\tfrac12\)
• \(g(3) = 3(3) + 2 = 11\)
• \(f(4) = 2(16) - 1 = 31\)
• \(g(5) = 3(5) + 2 = 17\)

\[\frac{f\!\left(-\tfrac12\right)\cdot g(3)}{f(4) - g(5)} = \frac{\left(-\tfrac12\right)(11)}{31 - 17} = \frac{-\tfrac{11}{2}}{14} = -\frac{11}{28}\]

(b) \(m \ast n = \dfrac{-n}{m^2 + 1}\). Test with \(m = -3,\, n = -10\):

\[(-3)\ast(-10) = \frac{-(-10)}{(-3)^2 + 1} = \frac{10}{10} = 1\]

\[(-10)\ast(-3) = \frac{-(-3)}{(-10)^2 + 1} = \frac{3}{101}\]

Since \(1 \neq \dfrac{3}{101}\), we have \((-3)\ast(-10) \neq (-10)\ast(-3)\). Therefore the operation \(\ast\) is not commutative.

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(a) Solve \((x-2)(x-3)=12\).

Expand the left side:

Bring every term to one side:

Factorise. Two numbers whose product is \(-6\) and whose sum is \(-5\) are \(-6\) and \(+1\):

Hence \(x-6=0\) or \(x+1=0\), giving:

(b) Area of the shaded (overlap) portion.

From the diagram the two circles have centre \(M\) and \(N\), each of radius \(r=7\text{ cm}\), and \(\angle PMQ=\angle PNQ=60^{\circ}\). The shaded lens is the region common to both circles. It is made up of two equal circular segments cut off by the common chord \(PQ\): one from circle \(M\) and one from circle \(N\).

Area of one segment = area of sector \(-\) area of triangle.

Area of sector \(PMQ\):

Area of triangle \(PMQ\) (two radii with included angle \(60^{\circ}\)):

Area of one segment:

The shaded lens is two such segments (one per circle):

Correct to the nearest whole number, the area of the shaded portion is \(9\text{ cm}^2\).

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Setup. The contracts are independent. Given:

\[P(\text{win }X) = 0.5,\qquad P(\text{not win }Y) = 0.3 \Rightarrow P(\text{win }Y) = 0.7\]

Also \(P(\text{not win }X) = 0.5\).

(a) Wins both contracts

\[P(X\cap Y) = 0.5\times 0.7 = 0.35\]

(b) Wins exactly one contract

Either win \(X\) and lose \(Y\), or lose \(X\) and win \(Y\):

\[P = (0.5\times 0.3) + (0.5\times 0.7) = 0.15 + 0.35 = 0.50\]

(c) Wins neither contract

\[P(\text{not }X\cap\text{not }Y) = 0.5\times 0.3 = 0.15\]

Check: \(0.35 + 0.50 + 0.15 = 1.00\), confirming all outcomes are accounted for.

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(a) Solve \(\dfrac{3}{2p-\tfrac{1}{2}} = \dfrac{\tfrac{1}{3}}{\tfrac{1}{4}p + 1}\)

Cross multiply:

\[3\left(\tfrac{1}{4}p + 1\right) = \tfrac{1}{3}\left(2p - \tfrac{1}{2}\right)\]

Expand both sides:

\[\tfrac{3}{4}p + 3 = \tfrac{2}{3}p - \tfrac{1}{6}\]

Collect the \(p\) terms and constants:

\[\tfrac{3}{4}p - \tfrac{2}{3}p = -\tfrac{1}{6} - 3\]

\[\left(\tfrac{9-8}{12}\right)p = -\tfrac{19}{6} \;\Rightarrow\; \tfrac{1}{12}p = -\tfrac{19}{6}\]

\[p = -\tfrac{19}{6}\times 12 = -38\]

\(p = -38\).

(b) Actual percentage profit

The marked price of GH¢760.00 gives a 20% profit, so it is \(120\%\) of the cost price:

\[\text{Cost price} = \frac{760}{1.20} = \text{GH¢}633.33\]

Sold at a 5% discount off the marked price:

\[\text{Selling price} = 760\times 0.95 = \text{GH¢}722.00\]

Actual profit \(= 722.00 - 633.33 = \text{GH¢}88.67\).

\[\text{Percentage profit} = \frac{88.67}{633.33}\times 100 \approx 14\%\]

Actual percentage profit \(\approx 14\%\) (to 2 s.f.).

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None

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(a) Multiplication table modulo 11 (multiply, then take the remainder on division by 11).

For example \(5\otimes 5 = 25 = 2(11)+3 \equiv 3\), \(9\otimes 9 = 81 = 7(11)+4 \equiv 4\), \(10\otimes 10 = 100 = 9(11)+1 \equiv 1\).

(i) \((9\otimes 5)\otimes(10\otimes 10) = 1 \otimes 1 = \mathbf{1}\).

(ii)(1) \(10\otimes m = 2\): from the row for 10, \(10\otimes 9 = 2\), so the truth set is \(\{9\}\).

(ii)(2) \(n\otimes n = 4\): scanning the leading diagonal, only \(9\otimes 9 = 4\), so the truth set is \(\{9\}\).

(b) The fraction. In lowest terms it equals \(\tfrac{3}{4}\), so let it be \(\dfrac{3k}{4k}\).

Numerator doubled is 34 more than the denominator:

Check: \(2(51) = 102 = 68 + 34\). The fraction is \(\mathbf{\tfrac{51}{68}}\).

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(a) Simplify \(3\sqrt{75} - \sqrt{12} + \sqrt{108}\)

Reduce each surd to lowest terms by taking out perfect squares:

\[\sqrt{75} = \sqrt{25\times 3} = 5\sqrt{3} \;\Rightarrow\; 3\sqrt{75} = 15\sqrt{3}\]

\[\sqrt{12} = \sqrt{4\times 3} = 2\sqrt{3}\]

\[\sqrt{108} = \sqrt{36\times 3} = 6\sqrt{3}\]

Combine the like surds:

\[15\sqrt{3} - 2\sqrt{3} + 6\sqrt{3} = 19\sqrt{3}\]

(b) If \(124_{n} = 232_{five}\), find \(n\).

Convert the right side to base ten:

\[232_{five} = 2(5^{2}) + 3(5) + 2 = 50 + 15 + 2 = 67\]

Express the left side in base ten:

\[124_{n} = 1(n^{2}) + 2(n) + 4 = n^{2} + 2n + 4\]

Set them equal and solve:

\[n^{2} + 2n + 4 = 67 \;\Rightarrow\; n^{2} + 2n - 63 = 0\]

\[(n+9)(n-7) = 0\]

Since a base must be positive, \(n = 7\).

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(a) For \(y=2\sin x+1\), the completed table is:

\(x\)	\(0^\circ\)	\(30^\circ\)	\(60^\circ\)	\(90^\circ\)	\(120^\circ\)	\(150^\circ\)	\(180^\circ\)	\(210^\circ\)	\(240^\circ\)	\(270^\circ\)
\(y\)	1.0	2.0	2.7	3.0	2.7	2.0	1.0	0.0	-0.7	-1.0

For example, \(y(60^\circ)=2\sin60^\circ+1=2(0.866)+1\approx2.7\).

(b) The required graph, plotted using the stated scales, is shown below.

(c) If \(\sin x=\frac14\), then

\[y=2\sin x+1=2\left(\frac14\right)+1=1.5.\]

Draw the horizontal line \(y=1.5\) and read its two points of intersection with the curve. Hence,

\[\boxed{x\approx15^\circ\text{ or }168^\circ}.\]

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(a) Shaded segment of the circle.

From the diagram, \(O\) is the centre, \(OX=OY=r\), \(\angle XOY=90^\circ\), and the shaded part is the segment lying between chord \(XY\) and the arc \(XKY\).

\[\text{Shaded area}=\text{area of sector } XOY-\text{area of }\triangle XOY.\]

Sector \(XOY\) subtends \(90^\circ\), so it is a quarter of the circle:

\[\text{sector}=\frac{90}{360}\pi r^{2}=\frac{1}{4}\pi r^{2}.\]

Triangle \(XOY\) is right-angled at \(O\) with legs \(r\) and \(r\):

\[\triangle XOY=\frac{1}{2}\,r\times r=\frac{1}{2}r^{2}.\]

Hence

\[504=\frac{1}{4}\pi r^{2}-\frac{1}{2}r^{2}=r^{2}\left(\frac{1}{4}\times\frac{22}{7}-\frac{1}{2}\right)=r^{2}\left(\frac{22}{28}-\frac{14}{28}\right)=r^{2}\left(\frac{8}{28}\right)=\frac{2}{7}r^{2}.\]

\[r^{2}=504\times\frac{7}{2}=1764\quad\Rightarrow\quad r=\sqrt{1764}=\boxed{42\text{ cm}}.\]

(b) Two isosceles triangles on a common base PQ.

Triangles \(PQR\) and \(PQS\) stand on opposite sides of the common base \(PQ\).

In \(\triangle PQR\) (isosceles with \(RP=RQ\)), the base angles are equal, and \(\angle PQR=66^\circ\) is a base angle, so

\[\angle RPQ=\angle RQP=66^\circ.\]

In \(\triangle PQS\) (isosceles with \(SP=SQ\)), \(\angle PSQ=109^\circ\) is the apex angle, so the base angles are

\[\angle SPQ=\angle SQP=\frac{180^\circ-109^\circ}{2}=\frac{71^\circ}{2}=35.5^\circ.\]

Since \(R\) and \(S\) are on opposite sides of \(PQ\), the angle \(RQS\) is the sum of the two base angles at \(Q\):

\[\angle RQS=\angle RQP+\angle PQS=66^\circ+35.5^\circ=\boxed{101.5^\circ}.\]

Maelezo ya Majibu

(a) Solving \(5\cos(x+8.5)^\circ - 1 = 0\).

\[\cos(x+8.5)^\circ = \frac{1}{5} = 0.2\]

\[x + 8.5 = \cos^{-1}(0.2) = 78.46^\circ\]

\[x = 78.46 - 8.5 = 69.96^\circ \approx 70^\circ\]

(b) Reading the bearings: the bearing of \(Q\) from \(P\) is \(150^\circ\) and the bearing of \(P\) from \(R\) is \(015^\circ\). Since the bearing of \(P\) from \(R\) is \(015^\circ\), the bearing of \(R\) from \(P\) is \(015^\circ + 180^\circ = 195^\circ\). With \(|PQ| = 24\) km and \(|PR| = 32\) km:

(i) Diagram. At \(P\), draw the north line. \(Q\) lies on a bearing of \(150^\circ\) at 24 km; \(R\) lies on a bearing of \(195^\circ\) at 32 km. The angle \(\angle QPR = 195^\circ - 150^\circ = 45^\circ\).

(ii) Distance \(|QR|\). By the cosine rule in \(\triangle PQR\):

\[|QR|^2 = 24^2 + 32^2 - 2(24)(32)\cos 45^\circ\]

\[= 576 + 1024 - 1536(0.7071) = 1600 - 1086.11 = 513.89\]

\[|QR| = \sqrt{513.89} = 22.67\text{ km (2 d.p.)}\]

(iii) Bearing of \(R\) from \(Q\). Taking \(P\) as origin with north as the \(y\)-axis: \(Q = (24\sin150^\circ,\,24\cos150^\circ) = (12,\,-20.78)\) and \(R = (32\sin195^\circ,\,32\cos195^\circ) = (-8.28,\,-30.91)\).

Displacement from \(Q\) to \(R\): \((-20.28,\,-10.13)\), which points south and west (third quadrant).

\[\theta = \tan^{-1}\!\left(\frac{20.28}{10.13}\right) = 63.5^\circ \text{ west of south}\]

\[\text{Bearing of }R\text{ from }Q = 180^\circ + 63.5^\circ = 243.5^\circ \approx 243^\circ\]

Maelezo ya Majibu

(a)(i) Adding all the regions of the Venn diagram and equating to \( n(U) = 125 \):

\[ (16 - 2x) + (6 + x) + (19 - 3x) + 5x + 7x + 8x + 4x + 4 = 125 \]

\[ 45 + 20x = 125 \implies 20x = 80 \implies x = 4 \]

(a)(ii) \( n(P \cup Q \cap R') = (16 - 2x) + 5x + (6 + x) = 22 + 4x \)

\[ = 22 + 4(4) = 22 + 16 = \mathbf{38} \]

(b) XW is a diameter, so \( \angle XYW = 90^\circ \) (angle in a semicircle).

In triangle XYW: \( \angle XWY + 90^\circ + 50^\circ = 180^\circ \), so \( \angle XWY = 40^\circ \).

(i) Since \( WX \parallel YZ \), \( \angle WYZ = \angle XWY = \mathbf{40^\circ} \) (alternate angles).

(ii) \( \angle WOZ = 2 \times \angle WYZ = 2 \times 40^\circ = 80^\circ \) (angle at centre is twice angle at circumference). In triangle WEO, \( \angle WEO = 180^\circ - (40^\circ + 80^\circ) = 60^\circ \), therefore \( \angle YEZ = \mathbf{60^\circ} \) (vertically opposite angles).

Maelezo ya Majibu

None

Maelezo ya Majibu

(a) Solve \(\dfrac{1}{x}+\dfrac{1}{y}=5\) and \(\dfrac{1}{y}-\dfrac{1}{x}=1\)

Let \(u=\dfrac{1}{x}\) and \(v=\dfrac{1}{y}\). Then

\[u+v = 5 \quad\text{and}\quad v-u = 1\]

Add the two equations:

\[2v = 6 \;\Rightarrow\; v = 3 \;\Rightarrow\; \frac{1}{y}=3 \;\Rightarrow\; y = \frac{1}{3}\]

Then \(u = 5 - v = 2\), so \(\dfrac{1}{x}=2 \Rightarrow x = \dfrac{1}{2}\).

\(x = \tfrac{1}{2},\; y = \tfrac{1}{3}\).

(b) Length of good surface

Total distance \(48\) km in \(45\) minutes \(= 0.75\) h. Let good surface \(= g\) km at \(72\) km/h and bad surface \(= b\) km at \(48\) km/h.

Distance: \(g + b = 48\).

Time: \(\dfrac{g}{72} + \dfrac{b}{48} = 0.75\). Multiply through by \(144\):

\[2g + 3b = 108\]

Substitute \(b = 48 - g\):

\[2g + 3(48 - g) = 108 \;\Rightarrow\; 2g + 144 - 3g = 108 \;\Rightarrow\; -g = -36\]

\[g = 36\]

The good surface is 36 km.